Question: Simplify the following expression and state the condition under which the simplification is valid. $p = \dfrac{6n^3 - 18n^2 - 240n}{n^3 - 7n^2 - 8n}$
Solution: First factor out the greatest common factors in the numerator and in the denominator. $ p = \dfrac {6n(n^2 - 3n - 40)} {n(n^2 - 7n - 8)} $ $ p = \dfrac{6n}{n} \cdot \dfrac{n^2 - 3n - 40}{n^2 - 7n - 8} $ Simplify: $ p = 6 \cdot \dfrac{n^2 - 3n - 40}{n^2 - 7n - 8}$ Since we are dividing by $n$ , we must remember that $n \neq 0$ Next factor the numerator and denominator. $ p = 6 \cdot \dfrac{(n - 8)(n + 5)}{(n - 8)(n + 1)}$ Assuming $n \neq 8$ , we can cancel the $n - 8$ $ p = 6 \cdot \dfrac{n + 5}{n + 1}$ Therefore: $ p = \dfrac{ 6(n + 5)}{ n + 1 }$, $n \neq 8$, $n \neq 0$